6.9E: Mazoezi ya Sehemu ya 6.9
- Page ID
- 178270
1) [T] Kupata maneno kwa\(\cosh x+\sinh x\) na\(\cosh x−\sinh x.\) kutumia calculator kwa grafu kazi hizi na kuhakikisha kujieleza yako ni sahihi.
- Jibu
- \(e^x\)na\(e^{−x}\)
2) Kutoka kwa ufafanuzi wa\(\cosh(x)\) na\(\sinh(x)\), pata antiderivatives yao.
3) Onyesha hilo\(\cosh(x)\) na\(\sinh(x)\) kukidhi\( y''=y\).
- Jibu
- Majibu inaweza kutofautiana
4) Tumia utawala wa quotient ili uhakikishe kuwa\(\dfrac{d}{dx}\big(\tanh(x)\big)=\text{sech}^2(x).\)
5) Pata\(\cosh^2(x)+\sinh^2(x)=\cosh(2x)\) kutoka kwa ufafanuzi.
- Jibu
- Majibu inaweza kutofautiana
6) Chukua derivative ya kujieleza uliopita ili kupata maelezo ya\(\sinh(2x)\).
7) Thibitisha\(\sinh(x+y)=\sinh(x)\cosh(y)+\cosh(x)\sinh(y)\) kwa kubadilisha maneno kwa maonyesho.
- Jibu
- Majibu inaweza kutofautiana
8) Chukua derivative ya kujieleza uliopita ili kupata maelezo\(\cosh(x+y).\)
Katika mazoezi 9 - 18, pata derivatives ya kazi zilizopewa na grafu pamoja na kazi ili kuhakikisha jibu lako ni sahihi.
9) [T]\(\cosh(3x+1)\)
- Jibu
- \(3\sinh(3x+1)\)
10) [T]\(\sinh(x^2)\)
11) [T]\(\dfrac{1}{\cosh(x)}\)
- Jibu
- \(−\tanh(x)\text{sech}(x)\)
12) [T]\(\sinh(\ln(x))\)
13) [T]\(\cosh^2(x)+\sinh^2(x)\)
- Jibu
- \(4\cosh(x)\sinh(x)\)
14) [T]\(\cosh^2(x)−\sinh^2(x)\)
15) [T]\(\tanh(\sqrt{x^2+1})\)
- Jibu
- \(\dfrac{x\text{sech}^2(\sqrt{x^2+1})}{\sqrt{x^2+1}}\)
16) [T]\(\dfrac{1+\tanh(x)}{1−\tanh(x)}\)
17) [T]\(\sinh^6(x)\)
- Jibu
- \(6\sinh^5(x)\cosh(x)\)
18) [T]\(\ln(\text{sech}(x)+\tanh(x))\)
Katika mazoezi 19 - 28, pata antiderivatives kwa kazi zilizopewa.
19)\(\cosh(2x+1)\)
- Jibu
- \(\frac{1}{2}\sinh(2x+1)+C\)
20)\(\tanh(3x+2)\)
21)\(x\cosh(x^2)\)
- Jibu
- \(\frac{1}{2}\sinh^2(x^2)+C\)
22)\(3x^3\tanh(x^4)\)
23)\(\cosh^2(x)\sinh(x)\)
- Jibu
- \(\frac{1}{3}\cosh^3(x)+C\)
24)\(\tanh^2(x)\text{sech}^2(x)\)
25)\(\dfrac{\sinh(x)}{1+\cosh(x)}\)
- Jibu
- \(\ln(1+\cosh(x))+C\)
26)\(\coth(x)\)
27)\(\cosh(x)+\sinh(x)\)
- Jibu
- \(\cosh(x)+\sinh(x)+C\)
28)\((\cosh(x)+\sinh(x))^n\)
Katika mazoezi 29 - 35, tafuta derivatives kwa kazi.
29)\(\tanh^{−1}(4x)\)
- Jibu
- \(\dfrac{4}{1−16x^2}\)
30)\(\sinh^{−1}(x^2)\)
31)\(\sinh^{−1}(\cosh(x))\)
- Jibu
- \(\dfrac{\sinh(x)}{\sqrt{\cosh^2(x)+1}}\)
32)\(\cosh^{−1}(x^3)\)
33)\(\tanh^{−1}(\cos(x))\)
- Jibu
- \(−\csc(x)\)
34)\(e^{\sinh^{−1}(x)}\)
35)\(\ln(\tanh^{−1}(x))\)
- Jibu
- \(−\dfrac{1}{(x^2−1)\tanh^{−1}(x)}\)
Katika mazoezi 36 - 42, tafuta antiderivatives kwa kazi.
36)\(\displaystyle ∫\frac{dx}{4−x^2}\)
37)\(\displaystyle ∫\frac{dx}{a^2−x^2}\)
- Jibu
- \(\dfrac{1}{a}\tanh^{−1}\left(\dfrac{x}{a}\right)+C\)
38)\(\displaystyle ∫\frac{dx}{\sqrt{x^2+1}}\)
39)\(\displaystyle ∫\frac{xdx}{\sqrt{x^2+1}}\)
- Jibu
- \(\sqrt{x^2+1}+C\)
40)\(\displaystyle ∫−\frac{dx}{x\sqrt{1−x^2}}\)
41)\(\displaystyle ∫\frac{e^x}{\sqrt{e^{2x}−1}}\)
- Jibu
- \(\cosh^{−1}(e^x)+C\)
42)\(\displaystyle ∫−\frac{2x}{x^4−1}\)
Katika mazoezi 43 - 45, tumia ukweli kwamba mwili unaoanguka na msuguano sawa na kasi ya mraba hutii equation\(\dfrac{dv}{dt}=g−v^2\).
43) Onyesha kwamba\(v(t)=\sqrt{g}\tanh(\sqrt{g}t)\) satisfies equation hii.
- Jibu
- Majibu inaweza kutofautiana
44) Pata usemi uliopita\(v(t)\) kwa kuunganisha\(\dfrac{dv}{g−v^2}=dt\).
45) [T] Tathmini jinsi mbali mwili imeshuka katika\(12\) sekunde kwa kutafuta eneo chini ya Curve ya\(v(t)\).
- Jibu
- \(37.30\)
Katika mazoezi 46 - 48, tumia hali hii: Cable kunyongwa chini ya uzito wake ina mteremko\(S=\dfrac{dy}{dx}\) unaotimiza\(\dfrac{dS}{dx}=c\sqrt{1+S^2}\). Mara kwa mara\(c\) ni uwiano wa wiani wa cable kwa mvutano.
46) Onyesha kwamba\(S=\sinh(cx)\) satisfies equation hii.
47)\(\dfrac{dy}{dx}=\sinh(cx)\) Unganisha ili kupata urefu cable\(y(x)\) kama\(y(0)=1/c\).
- Jibu
- \(y=\frac{1}{c}\cosh(cx)\)
48) Mchoro cable na kuamua jinsi mbali chini sags katika\(x=0\).
Katika mazoezi 49 - 52, tatua kila tatizo.
49) [T] mlolongo hangs kutoka posts mbili\(2\) m mbali na kuunda catenary ilivyoelezwa na equation\(y=2\cosh(x/2)−1\). Pata mteremko wa catenary kwenye chapisho la uzio wa kushoto.
- Jibu
- \(−0.521095\)
50) [T] mlolongo hangs kutoka posts mbili mita nne mbali na kuunda catenary ilivyoelezwa na equation\(y=4\cosh(x/4)−3.\) Kupata urefu jumla ya catenary (safu urefu).
51) [T] Mstari wa nguvu ya juu-voltage ni catenary iliyoelezwa na\(y=10\cosh(x/10)\). Pata uwiano wa eneo chini ya catenary kwa urefu wake wa arc. Unaona nini?
- Jibu
- \(10\)
52) Mstari wa simu ni catenary iliyoelezwa na\(y=a\cosh(x/a).\) Pata uwiano wa eneo chini ya catenary kwa urefu wake wa arc. Je, hii inathibitisha jibu lako kwa swali la awali?
53) Thibitisha formula kwa derivative ya\(y=\sinh^{−1}(x)\) kwa kutofautisha\(x=\sinh(y).\)
(Kidokezo: Tumia utambulisho wa trigonometric ya hyperbolic.)
54) Thibitisha formula kwa derivative ya\(y=\cosh^{−1}(x)\) kwa kutofautisha\(x=\cosh(y).\)
(Kidokezo: Tumia utambulisho wa trigonometric ya hyperbolic.)
55) Thibitisha formula kwa derivative ya\(y=\text{sech}^{−1}(x)\) kwa kutofautisha\(x=\text{sech}(y).\)
(Kidokezo: Tumia utambulisho wa trigonometric ya hyperbolic.)
56) Thibitisha kwamba\(\cosh(x)+\sinh(x))^n=\cosh(nx)+\sinh(nx).\)
57) Thibitisha usemi\(x=\sinh(y)=\dfrac{e^y−e^{−y}}{2}\) kwa\(\sinh^{−1}(x).\) Kuzidisha\(2e^y\) na kutatua kwa\(y\). Je! Maneno yako yanafanana na kitabu?
58) Thibitisha usemi\(x=\cosh(y)=\dfrac{e^y+e^{−y}}{2}\) kwa\(\cosh^{−1}(x).\) Kuzidisha\(2e^y\) na kutatua kwa\(y\). Je! Maneno yako yanafanana na kitabu?